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Pump Power Calculator

Calculate hydraulic power, shaft power, motor power, specific speed, and annual energy cost for pumps. Supports multiple fluid types and unit systems.

Hydraulics & Water Resources Mechanical engineers, civil engineers, MEP designers, plant engineers Commercial Intent: MEDIUM
75%
90%

Engineering Formulas

Hydraulic Power

Ph = ρ × g × Q × H Ph (kW) = ρ × g × Q × H / 1000 Q in m³/s, ρ in kg/m³, H in m
P_h: Hydraulic power (kW)
ρ: Fluid density (kg/m³)
g: Gravity (9.81 m/s²)
Q: Flow rate (m³/s)
H: Total head (m)

Shaft & Motor Power

Ps = Ph / η Pm = Ps / ηm η = pump efficiency, ηm = motor efficiency
P_s: Shaft power (kW)
P_m: Motor input power (kW)
η: Pump efficiency (decimal)
η_m: Motor efficiency (decimal)

Specific Speed

Ns = N × √Q / H0.75 Radial: Ns < 80 Mixed: 80 < Ns < 170 Axial: Ns > 170
N_s: Specific speed
N: Motor speed (RPM)
Q: Flow rate (m³/s at best efficiency)
H: Head per stage (m)

Annual Energy Cost

Annual Cost = Pm × Hours × Rate Annual MWh = Pm × Hours / 1000
P_m: Motor power (kW)
Hours: Annual operating hours
Rate: Energy rate ($/kWh)

Worked Example

Water Pump for Building Supply

flowRate: 50flowUnit: L/shead: 30headUnit: mfluidType: waterpumpEfficiency: 75motorEfficiency: 90motorRPM: 1450annualHours: 4000energyRate: 0.12
Flow Conversion
Q = 50 L/s = 0.05 m³/s
Hydraulic Power
Ph = 1000 × 9.81 × 0.05 × 30 / 1000 = 14.72 kW (19.7 HP)
Shaft Power
Ps = 14.72 / 0.75 = 19.62 kW (26.3 HP)
Motor Power
Pm = 19.62 / 0.90 = 21.80 kW (29.2 HP)
Specific Speed
Ns = 1450 × √0.05 / 30^0.75 = 25.3 → Radial pump
Annual Cost
21.80 × 4000 × 0.12 = $10,464/yr
Result: 21.8 kW motor, radial pump, $10,464 annual energy cost

Engineering Notes

Always select pump to operate near BEP (best efficiency point).
For variable flow systems, consider VFDs to reduce energy costs.
NPSH available must exceed NPSH required to avoid cavitation.
Parallel pumps: total flow increases but not linearly.
Series pumps: total head adds up — used for high-head applications.

Assumptions

• Steady flow conditions
• Constant density (incompressible flow)
• Single-stage pump
• No cavitation
• Standard gravity g = 9.81 m/s²

Common Mistakes

✕ Using flow rate in L/s without converting to m³/s
✕ Forgetting to include motor efficiency in power calculation
✕ Using head in ft without converting to meters
✕ Selecting wrong pump type for specific speed range
✕ Not accounting for fluid density differences

Frequently Asked Questions

What is the difference between hydraulic, shaft, and motor power?

Hydraulic power is the power transferred to the fluid. Shaft power includes pump losses. Motor power is the electrical input, including motor losses. Each stage adds inefficiency.

What is specific speed and why is it important?

Specific speed characterizes pump geometry and flow type. Low Ns = radial (high head, low flow), high Ns = axial (low head, high flow). It helps select the right pump type.

What pump efficiency should I expect?

Centrifugal pumps: 50-85%. Mixed flow: 70-85%. Axial flow: 65-80%. Small pumps (<10 kW) tend toward lower efficiency, large pumps (>100 kW) toward higher.

How do I reduce pump energy costs?

Use VFDs for variable flow, select pumps at BEP (best efficiency point), use larger diameter pipes to reduce head loss, and maintain pumps regularly.

What fluid density should I use for slurry?

Slurry density depends on solids concentration. Use 1100-1300 kg/m³ for typical slurries. For precise values, calculate as: ρ = ρ_liquid × (1 - C_v) + ρ_solid × C_v.

References & Standards

Hydraulic Institute StandardsISO 9906
Hydraulic Institute Standards
Pump design, selection, and testing standards
ISO 9906
Rotodynamic pumps — Hydraulic performance acceptance tests
Karassik et al.
Pump Handbook — comprehensive pump reference
ANSI/HI 1.3
Rotodynamic Centrifugal Pumps for Design and Application

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